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3 years ago

Solve the inequality: 4(x-3)-7<2(x-3)

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

5 0

Answer:

$x$ $\frac{13}{2}$

Step-by-step explanation:

$4x-12-7$

$4x-19$

$4x$

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If 45.78 ÷ 1.7 is written in long division form as long division set up where 17 is on the outside of the division symbol and 45

saveliy_v [14]

Answer:

The decimal point should be placed after the 0 in 45780.

Step-by-step explanation:

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3 years ago

Can someone pls help me

MAVERICK [17]

Standard deviation I think

6 0

3 years ago

Find the approximate perimeter of ABC plotted below.

maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}$

$AB = \sqrt{4 + 4}$

$AB = \sqrt{16}$

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}$

$BC = \sqrt{25 + 49}$

$BC = \sqrt{74}$

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}$

$AC = \sqrt{47 + 25}$

$AC = \sqrt{74}$

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 0

3 years ago

The answer that go in the box? Please help

saul85 [17]

Transaction 3 - $143
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2 years ago

Which of the following is a point slope equation of the line below (10,5)(2,2)

salantis [7]

Point slope form follows the equation y-y₁=m(x-x₁), so we want it to look like that. Starting off with m, or the slope, we can find this using your two points with the formula $\frac{ y_{1} -y_{2} }{ x_{1} - x_{2} }$ . Note that y₁ and x₁ are from the same point, but it does not matter which point you designate to be point 1 and point 2. Thus, we can plug our numbers in - the x value comes first in the equation, and the y value comes second, so we have
$\frac{5-2}{10-2} = \frac{3}{5}$ as our slope. Keeping in mind that it does not matter which point is point 1 and which point is point 2, we go back to y-y₁=m(x-x₁) and plug a point in (I'll be using (10,5)). Note that x₁, m, and y₁ need to be plugged in, but x and y stay that way so that you can plug x or y values into the formula to find where exactly it is on the line. Thus, we have our point slope equation to be $y- 5= \frac{3}{5} (x-10)$

Feel free to ask further questions!

7 0

3 years ago